If the work done on the system or by the system is zero, which one of the following statements for a gas kept at a certain temperature is correct?
- (a)Change in internal energy of the system is equal to flow of heat in or out of the system.
- (b)Change in internal energy of the system is less than heat transferred.
- (c)Change in internal energy of the system is more than the heat flow.
- (d)Cannot be determined.
Answer
Why
Correct - A, the change in internal energy equals the heat flowing in or out. The first law of thermodynamics is Delta U = Q - W, where Q is the heat added to the system and W is the work done BY the system. If no work is done (W = 0), the equation reduces to Delta U = Q: every joule of heat entering or leaving the gas goes entirely into changing its internal energy. So the change in internal energy is exactly equal to the heat flow.
Why the others are wrong
- (b)Change in internal energy of the system is less than heat transferred. — For the internal-energy change to be less than the heat transferred, the gas would have to do positive work (W greater than 0); but the question fixes W = 0, so no heat is diverted into work.
- (c)Change in internal energy of the system is more than the heat flow. — For the change to exceed the heat flow, work would have to be done ON the gas to add extra internal energy, again contradicting W = 0.
- (d)Cannot be determined. — With W = 0 the first law gives a definite result, Delta U = Q, so the change is fully determined, not indeterminate.
Concept
The first law of thermodynamics is a statement of energy conservation for a gas: the change in internal energy equals the heat added minus the work done by the gas (Delta U = Q - W). Internal energy is a state property, depending for an ideal gas only on temperature. When the process does zero work, the heat exchanged and the internal-energy change are numerically identical.
Identify what is zero. Setting the work done on or by the system to zero removes the W term and leaves Delta U = Q. This is the constant-volume (isochoric) case, where all the heat changes the internal energy.
Key facts
- The first law states Delta U = Q - W, with Q the heat added and W the work done by the system.
- If W = 0, then Delta U = Q.
- Zero work typically corresponds to a constant-volume (isochoric) process.
- For an ideal gas, internal energy depends only on temperature.
With W = 0 the first law reduces to Delta U = Q - option (a).
Study next
Common traps
- Getting the sign of W wrong - here W is the work done BY the gas.
- Forgetting that a constant-volume process does zero work, so all heat becomes internal energy.
Asked by applying the first law to a process where either the work or the heat is zero, or by identifying the constant-volume case.
Related PYQs
No directly related past PYQ was found.
Practice
- practice — not a real PYQ
In a constant-volume process, the heat supplied to a gas goes entirely into
- (a)work done by the gas
- (b)change in internal energy
- (c)change in pressure only
- (d)latent heat
Answer(b) change in internal energy - no work is done at constant volume. - practice — not a real PYQ
The first law of thermodynamics is a statement of the conservation of
- (a)charge
- (b)momentum
- (c)energy
- (d)mass
Answer(c) energy - it balances heat, work and internal energy.